# Transported measures

The following theorem is a smoothness result for measures trasported onto smaller dimensional spaces. What’s interesting about this technical result is a decomposition needed in the proof which is rather general in principle and useful in that it doesn’t require much knowledge of the geometry of the problem.

Here, suppose ${\Phi:\overline{B_m}\rightarrow \mathbb{R}^n}$ is a ${C^\infty}$ map from the closed unit ball ${\overline{B_m}\subset \mathbb{R}^m}$ in ${\mathbb{R}^n}$, where ${m\geq n}$. Let ${\psi \in C^{(k+1)}\left(\overline{B_m}\right)}$ with compact support in ${B_m}$. Then we can transport measure ${\psi(\tau)d\tau}$ from ${B_m}$ to ${\mathbb{R}^n}$ via ${\Phi}$, i.e.

$\displaystyle d\mu=\Phi_\ast (\psi(\tau)d\tau),$

defined by

$\displaystyle \int_{\mathbb{R}^n}{f(x)}\,d\mu(x)=\int_{B_m}{f(\Phi(\tau))\psi(\tau)}\,d\tau.$

If ${\Phi}$ is a well behaved map we shall prove that the transported measure is indeed quite regular:

Theorem 1 Let ${\Phi}$ be also in ${C^{(k+2)}(\overline{B_m})}$. Let ${J}$ be the determinant of a ${n\times n}$ submatrix of the Jacobian of ${\Phi}$, and suppose that there exists a multiindex ${\alpha}$ such that

$\displaystyle |\partial^{\alpha}_{\tau}J(\tau)|>0$

everywhere on the ball ${\overline{B}_m}$. Then the transported measure ${d\mu=\Phi_\ast (\psi(\tau)d\tau)}$ is absolutely continuous with respect to the Lebesgue measure on ${\mathbb{R}^n}$. Moreover, its density ${\rho}$ has the ${L^1}$-Hölder smoothness property

$\displaystyle \int_{\mathbb{R}^n}{|\rho(x+t)-\rho(x)|}\,dx\leq C \|t\|^{\delta}$

for every ${t\in\mathbb{R}^n}$, where the exponent ${\delta}$ can be choosen in the range ${0<\delta <(2|\alpha|)^{-1}}$. (The constant ${C}$ will depend only on the ${C^{(k+2)}(\overline{B_m})}$-norm of ${\Phi}$, the ${C^1}$-norm of ${\psi}$, a lower bound for ${|\partial^{\alpha}_{\tau}J(\tau)|}$, ${\delta}$ and ${|\alpha|}$).

Before addressing the proof of this theorem we state and prove a lemma which is of interest on his own and on which proof we’ll digress a little.

Lemma 2 Let ${F}$ be a function in ${C^{(k+2)}(\overline{B})}$ and suppose there exists an ${\alpha}$ and a ${b>0}$ s.t.

$\displaystyle |\partial^{\alpha}_{\tau}F(\tau)|>b$

on all of ${\overline{B}}$. Then one has the finiteness of the following integral

$\displaystyle \int_{\overline{B}}{|F(\tau)|^{-\sigma}}\,d\tau \leq A < \infty$

for every ${0<\sigma < |\alpha|^{-1}}$, and the upperbound ${A}$ only depends on ${\|F\|_{C^{(k+2)}(\overline{B_m})}, b, \sigma, |\alpha|}$ and the volume of the ball.

Proof: One has, for ${0<\sigma <1}$,

$\displaystyle |y|^{-\sigma}=c_\sigma \int_{-\infty}^{\infty}{e^{2\pi i \lambda y}|\lambda|^{\sigma-1}}\,d\lambda.$

To see this, notice it’s just the Fourier transform of ${|\lambda|^{\sigma-1}}$, which is an even function, and by scaling invariance one has

$\displaystyle \widehat{|\lambda|^{\sigma-1}}(y)=\int_{-\infty}^{\infty}{e^{2\pi i \beta\lambda y}|\beta\lambda|^{\sigma-1}}\,\beta d\lambda= \beta^{\sigma}\widehat{|\lambda|^{\sigma-1}}(\beta y),$

and it suffices to choose ${\beta=|y|^{-1}}$.

Now, using this fact we have

$\displaystyle \int_{\overline{B}}{|F(\tau)|^{-\sigma}}\,d\tau = \int_{\overline{B}}{\int_{-\infty}^{\infty}{e^{2\pi i \lambda F(\tau)}|\lambda|^{\sigma-1}}\,d\lambda}\,d\tau,$

and by Fubini this is dominated in absolute value by

$\displaystyle \int_{-\infty}^{\infty}{|\lambda|^{\sigma-1}\left|\int_{B}{e^{2\pi i \lambda F(\tau)}}\,d\tau\right|}\,d\lambda.$

Now the lower boundedness of a derivative of ${F}$ comes into play as we invoke Van der Corput’s lemma to estimate the decay of the inner integral. We have

$\displaystyle \left|\int_{B}{e^{2\pi i \lambda F(\tau)}}\,d\tau\right| \lesssim \min\{1,(b\lambda)^{-\frac{1}{|\alpha|}}\},$

(notice that since this is a multidimensional integral the constant depends on the size of the ball ${B}$ as well) and since we allow our constants to depend on ${b}$ and ${|\alpha|}$ as well, we have that

$\displaystyle \int_{-\infty}^{\infty}{|\lambda|^{\sigma-1}\min\{1,|\lambda|^{-\frac{1}{|\alpha|}}\}}\,d\lambda$

is finite as long as ${\sigma-1-\frac{1}{|\alpha|}<-1}$. $\Box$

The proof of the lemma ends here but we can say more actually. We don’t explicitely need to call into play Var der Corput’s lemma as what we really want to know is an estimate on the size of ${F}$. Now, it is generally true that given a function ${f}$ one has

$\displaystyle \|f\|_{L^p}^p\sim_p \sum_{j\in\mathbb{Z}}{2^{-jp} \left|\{x\,:\,|f(x)|\sim 2^{-j}\}\right|}, \ \ \ \ \ (1)$

and this is true even when ${p<1}$, so that what we really want is an estimate on the size of the sublevel sets of ${F}$. To achieve this, one can use the following lemma, whose proof follows the same lines as the one of Van der Corput’s lemma (see this other post):

Lemma 3 Let ${f}$ be a function in ${C^k(\overline{B})}$ such that ${|\partial^{\alpha}f(x)|\geq 1}$, then one has

$\displaystyle \left|\{x\,:\, |f|\leq \lambda\}\right| \lesssim \lambda^{\frac{1}{|\alpha|}}.$

Notice the constant depends on ${u}$ as well as ${|\alpha|}$.

Inserting this into (1) – remember ${F}$ is bounded from above by its ${C^{(k+2)}}$-norm – we have the integral is dominated by

$\displaystyle \lesssim\sum_{j\geq j_0}{2^{j\sigma} 2^{-\frac{j}{|\alpha|}}},$

which is finite if ${\sigma < |\alpha|^{-1}}$.

We now come to the proof of the regularity properties of the transported measure. Proof: By lemma 2 we see that the zero set ${Z:=\{\tau\in\overline{B}\,:\,J(\tau)=0\}}$ has zero measure. We’ll cover ${B\backslash Z}$ by a countable union of balls such that on each ball ${B_j}$ the value of ${J}$ is roughly constant and actually ${\sim r_j}$ (the radius), and that the dilated balls are disjoint from ${Z}$ and have the bounded overlapping property.

In more detail, fix ${\varepsilon \ll 1}$ and to every point ${\tau\in B\backslash Z}$ associate the ball ${B(\tau,c |J(\tau)|)}$, where the constant ${c}$ is chosen small enough to satisfy the following: for all ${\tau' \in B(\tau,8c |J(\tau)|)}$ one has

$\displaystyle 1-\varepsilon \leq \left|\frac{J(\tau')}{J(\tau)}\right|\leq 1+\varepsilon.$

The existence of such a constant is guaranteed by the fact that the first derivative of ${J}$ is bounded (i.e. Lipschitz), ${\Phi}$ being of finite ${C^{(k+2)}}$-norm. So now we have fixed the constant ${c}$ such that on each ball ${J}$ is roughly constant, and we proceed to take a maximal disjoint subfamily of balls whose centers are ${\{\tau_j\}_{j\in\mathbb{N}}}$. We dilate the balls of a factor ${4}$, to get the balls ${B_j:=B(\tau_j,r_j)}$, where ${r_j:=4c|J(\tau_j)|}$. We have to prove this is a covering of ${B\backslash Z}$: take any ${\tau \in B\backslash Z}$ and the associated ball ${B(\tau)=B(\tau,c|J(\tau)|)}$. Since we’ve chosen the subfamily to be maximal disjoint then there is an index ${k}$ such that ${B(\tau)\cap B(\tau_k, c|J(\tau_k)|)\neq \emptyset}$, so that there are two possible cases depending on which ball is bigger: either ${\tau\in B(\tau_k,3c|J(\tau_k)|)}$ or ${\tau_k\in B(\tau,3c|J(\tau)|)}$. In the first case we’re done. Anyway, in both cases one has by construction

$\displaystyle |J(\tau)|\leq \max\{1+\varepsilon,\frac{1}{1-\varepsilon}\}|J(\tau_k)|=\frac{1}{1-\varepsilon}|J(\tau_k)|,$

so that for ${\varepsilon}$ small enough

$\displaystyle |\tau_k-\tau|\leq 3c|J(\tau)|\leq 4c|J(\tau_k)|,$

that is ${\tau \in B_k}$.

Now, the balls ${B^{\ast}_j:=2 B_j}$ are clearly disjoint from ${Z}$ (the have radiuses ${2r_j=8c|J(\tau_j)|}$), and have the bounded overlapping property: balls that overlap in a point ${\tau}$ have comparable values of ${|J|}$ on them and therefore comparable radiuses, and balls ${\frac{1}{8}B^{\ast}_j}$ are disjoint by construction, so that there’s only a finite number of them that can fit in a sphere centered in ${\tau}$ and of radius comparable to ${|J(\tau)|}$.

So far we have got this nice covering where ${|J|\sim r}$, now we want to use it to decompose the measure ${d\mu}$ into pieces. Choose a partition of unity ${\{\eta_j\}_j}$ subordinated to the collection ${B_j}$ and such that ${\|\nabla \eta_j\|\lesssim r^{-1}_j}$. Define

$\displaystyle d\mu_j=\Phi_{\ast}(\psi(\tau)\eta_j(\tau)d\tau).$

Actually this is not exactly the decomposition we want. We know that ${J}$ is non zero on every ball, but that means ${\Phi}$ is locally invertible there. Then we want to use ${\Phi}$ as a change of variable to be able to find a density on each ball; summing the densities we’ll have a density for ${d\mu}$.

So, suppose wlog that the submatrix of ${d\Phi}$ whose determinant is ${J}$ is the one obtained choosing the first ${n}$ coordinates, so that we can write ${\tau=(x,y)}$ with ${x\in \mathbb{R}^n, y\in\mathbb{R}^{m-n}}$. Define

$\displaystyle \Phi_{j,y}=\Phi|_{B_j'\times \{y\}},$

where ${B_j'}$ is the projection of the ball ${B_j}$ onto the first ${n}$ coordinates, i.e. ${B_j'=B_j\cap \left(\mathbb{R}^n\times \{0\}^{m-n}\right)}$. Notice that ${\det d\Phi_{j,y}=J}$, and this means that ${\Phi_{j,y}}$ is actually a diffeomorphism between the first ${n}$ coordinates of ${\tau}$ and ${\mathbb{R}^n}$ (the range of ${\Phi}$). Then we define

$\displaystyle d\mu_{j,y}=\left(\Phi_{j,y}\right)_{\ast}(\psi(\cdot,y)\eta_j(\cdot,y) dx');$

we apply the change of variable we’ve been talking about and write explicitly the density of ${d\mu_{j,y}}$:

$\displaystyle \int_{\mathbb{R}^n}{f(x)}\,d\mu_{j,y}(x)=\int_{B}{f(\Phi_{j,y}(x'),y)\psi(x',y)\eta_j(x',y)}\,dx'$

$\displaystyle =\int_{B}{f(x)\psi(\Phi_{j,y}^{-1}(x),y)\eta_j(\Phi_{j,y}^{-1}(x),y)|J(\Phi_{j,y}^{-1}(x),y)|^{-1}}\,dx,$

so that ${d\mu_{j,y}(x)=\rho_{j,y}(x)dx}$ with

$\displaystyle \rho_{j,y}(x)=\psi(\Phi_{j,y}^{-1}(x),y)\eta_j(\Phi_{j,y}^{-1}(x),y)|J(\Phi_{j,y}^{-1}(x),y)|^{-1}. \ \ \ \ \ (2)$

What we’ll want at the end is the density

$\displaystyle \rho_j(x)=\int_{B_j''}{\rho_{j,y}(x)}\,dy,$

where ${B_j''}$ is the projection of the ball of index ${j}$ onto the last ${m-n}$ coordinates. Notice that ${B_j \subset B_j'\times B_j'' \subset B_j^{\ast}}$. We’ll get there in a moment.

We have two estimates, one for the ${L^1}$ norm of ${\rho_{j,y}}$ and one for the ${L^1}$ norm of ${\nabla \rho_{j,y}}$: since both ${\psi}$ and ${\eta_j}$ are bounded, and the last one restricts the integral to ${B_j'}$,

$\displaystyle \int_{\mathbb{R}^n}{|\rho_{j,y}(x)|}\,dx\lesssim \int_{\mathbb{R}^n}{|J(\Phi_{j,y}^{-1}(x),y)|^{-1}}dx=\int_{B_j'}{1}dx'\sim r_j^{n}.$

On the other hand one has

$\displaystyle \nabla_{x'}\left(\rho_{j,y}\left(\Phi_{j,y}(x')\right)\right)=\nabla \rho_{j,y}\left(\Phi_{j,y}(x')\right)d\Phi_{j,y}(x') \ \ \ \ \ (3)$

$\displaystyle = \nabla_{x'}\left(\psi(x',y)\eta_j(x',y)|J(x',y)|^{-1}\right)=-\frac{\nabla_{x'}J}{|J|^2}\psi\eta_j + |J|^{-1}\eta_j \nabla_{x'}\psi + |J|^{-1}\psi \nabla_{x'}\eta_j$

(I started suppressing some details in order not to freak out). Let’s look at those last terms and the contribution they have in absolute value. Remember ${|J|\sim r_j}$ on the ball and ${\Phi}$ has ${C^{(k+2)}}$-norm finite, therefore

$\displaystyle \left|\frac{\nabla_{x'}J}{|J|^2}\psi\eta_j\right|\lesssim r^{-2}_j.$

${\psi}$ has finite ${C^1}$-norm, therefore

$\displaystyle \left||J|^{-1}\eta_j \nabla_{x'}\psi\right|\lesssim r^{-1}_j,$

and by our choice of partition ${\eta}$

$\displaystyle \left||J|^{-1}\psi \nabla_{x'}\eta_j\right|\lesssim r^{-2}_j.$

Since the ${r_j}$ are bounded from above by the constant ${1}$, ${r_j^{-1}\leq r_j^{-2}}$ , and one has from (3) that

$\displaystyle \int_{\mathbb{R}^n}{\nabla \rho_{j,y}(x)}dx \lesssim r^{-2}_j\int_{\mathbb{R}^n}{|J(\Phi_{j,y}^{-1}(x),y)|^{-1}}dx\sim r_j^{n-2}.$

These two estimates imply the following estimates for our true densities ${\rho_j}$ (take the integral on ${B_j''}$):

$\displaystyle \int_{\mathbb{R}^n}{|\rho_j|}dx \lesssim r_j^{m},$

$\displaystyle \int_{\mathbb{R}^n}{|\nabla\rho_j|}dx \lesssim r_j^{m-2}.$

Now we can finally conclude.${\rho=\sum_j{\rho_j}}$ is the density of ${d\mu}$ and we want to show the finiteness of

$\displaystyle \sup_{t\in\mathbb{R}^n}{\|t\|^{-\delta}\int_{\mathbb{R}^n}{|\rho(x+t)-\rho(x)|}dx}$

under the condition that ${\delta}$ is small enough. By using our ${L^1}$ estimates on ${\rho_j}$ and ${\nabla\rho_j}$ we can prove

$\displaystyle \|t\|^{-\delta}\int_{\mathbb{R}^n}{|\rho(x+t)-\rho(x)|}dx \lesssim 2\|t\|^{-\delta}r_j^{m}$

and

$\displaystyle \|t\|^{-\delta}\int_{\mathbb{R}^n}{|\rho(x+t)-\rho(x)|}dx \lesssim \|t\|^{1-\delta}r_j^{m-2},$

the second one coming from a straightforward application of the intermediate value theorem. Then we optimize, since ${\|t\|^{-\delta}r_j^{m}\leq \|t\|^{1-\delta}r_j^{m-2}}$ iff ${r_j\leq \|t\|^{1/2}}$, we have

$\displaystyle \|t\|^{-\delta}\int_{\mathbb{R}^n}{|\rho(x+t)-\rho(x)|}dx \lesssim \sum_{j\,:\,r_j\leq \|t\|^{1/2}}{\|t\|^{-\delta}r_j^{m}}+\sum_{j\,:\,r_j> \|t\|^{1/2}}{\|t\|^{1-\delta}r_j^{m-2}},$

and each summand is dominated by ${r_j^{m-2\delta}}$. So we have found the bound ${\sum_{j}{r_j^{m-2\delta}}}$ and it only remains to be proven that this is indeed finite. But this is just the content of the lemma we proved at the beginning, since

$\displaystyle \sum_{j}{r_j^{m-2\delta}}\sim \int_{B}{|J|^{-2\delta}}d\tau.$

The contribution of the ball ${B_j}$ to the integral is ${\sim Vol(B_j)|J(\tau_j)|^{-2\delta}\sim r_j^{m} r_j^{-2\delta}}$, and the balls have the bounded overlapping property, so that in the sum every point is counted at most a fixed number of times. The sum is therefore finite when ${0<\delta < (2|\alpha|)^{-1}}$. $\Box$

See for reference:

• Singular and maximal Radon transforms, section 7 – Christ Nagel, Stein, Wainger (1999)
• Harmonic Analysis on Nilpotent Groups and Singular Integral II: singular kernels supported on submanifolds – Ricci,Stein (Journal of Functional Analysis, 1988)